Difference between revisions of "2003 AIME I Problems/Problem 13"
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The sum of the elements on or to the right of the line of symmetry is thus <math>\frac{2047 + 351}{2} = 1199</math>. However, we also counted the <math>44</math> numbers from <math>2004</math> to <math>2^{11}-1 = 2047</math>. Indeed, all of these numbers have at least <math>6</math> <math>1</math>'s in their base-<math>2</math> representation, as all of them are greater than <math>1984 = 11111000000_2</math>, which has <math>5</math> <math>1</math>'s. Therefore, our answer is <math>1199 - 44 = 1155</math>, and the remainder is <math>\boxed{155}</math>. | The sum of the elements on or to the right of the line of symmetry is thus <math>\frac{2047 + 351}{2} = 1199</math>. However, we also counted the <math>44</math> numbers from <math>2004</math> to <math>2^{11}-1 = 2047</math>. Indeed, all of these numbers have at least <math>6</math> <math>1</math>'s in their base-<math>2</math> representation, as all of them are greater than <math>1984 = 11111000000_2</math>, which has <math>5</math> <math>1</math>'s. Therefore, our answer is <math>1199 - 44 = 1155</math>, and the remainder is <math>\boxed{155}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We seek the number of allowed numbers which have <math> k</math> 1's, not including the leading 1, for <math> k=0, 1, 2, \ldots , 10</math>. | ||
+ | |||
+ | For <math> k=0,\ldots , 4</math>, this number is | ||
+ | |||
+ | <math> \binom{k}{k}+\binom{k+1}{k}+\cdots+\binom{2k}{k}</math>. | ||
+ | |||
+ | By the Hockey Stick Identity, this is equal to <math> \binom{2k+1}{k+1}</math>. So we get | ||
+ | |||
+ | <math> \binom{1}{1}+\binom{3}{2}+\binom{5}{3}+\binom{7}{4}+\binom{9}{5}=175</math>. | ||
+ | |||
+ | For <math> k=5,\ldots , 10</math>, we end on <math> \binom{10}{k}</math> - we don't want to consider numbers with more than 11 digits. So for each <math> k</math> we get | ||
+ | |||
+ | <math> \binom{k}{k}+\binom{k+1}{k}+\binom{10}{k}=\binom{11}{k+1}</math> | ||
+ | |||
+ | again by the Hockey Stick Identity. So we get | ||
+ | |||
+ | <math> \binom{11}{6}+\binom{11}{7}+\binom{11}{8}+\binom{11}{9}+\binom{11}{10}+\binom{11}{11}=\frac{2^{11}}{2}=2^{10}=1024</math>. | ||
+ | |||
+ | The total is <math> 1024+175=1199</math>. Subtracting out the <math> 44</math> numbers between <math> 2003</math> and <math> 2048</math> gives <math> 1155</math>. Thus the answer is <math> 155</math>. | ||
== See also == | == See also == |
Revision as of 09:31, 4 July 2018
Contents
Problem
Let be the number of positive integers that are less than or equal to
and whose base-
representation has more
's than
's. Find the remainder when
is divided by
.
Solution
In base- representation, all positive numbers have a leftmost digit of
. Thus there are
numbers that have
digits in base
notation, with
of the digits being
's.
In order for there to be more 's than
's, we must have
. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle, from rows
to
(as
). Since the sum of the elements of the
th row is
, it follows that the sum of all elements in rows
through
is
. The center elements are in the form
, so the sum of these elements is
.
The sum of the elements on or to the right of the line of symmetry is thus . However, we also counted the
numbers from
to
. Indeed, all of these numbers have at least
's in their base-
representation, as all of them are greater than
, which has
's. Therefore, our answer is
, and the remainder is
.
Solution 2
We seek the number of allowed numbers which have 1's, not including the leading 1, for
.
For , this number is
.
By the Hockey Stick Identity, this is equal to . So we get
.
For , we end on
- we don't want to consider numbers with more than 11 digits. So for each
we get
again by the Hockey Stick Identity. So we get
.
The total is . Subtracting out the
numbers between
and
gives
. Thus the answer is
.
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.